A big cruise ship dropped anchor off the Caribbean island of Antigua. The heavy anchor dropped into the water at a rate of $2.5$ meters per second. After $45$ seconds, the anchor was $40$ meters below the water's surface. From what height (above the water's surface) was the anchor released?
The anchor dropped at a rate of $2.5$ meters per second, so it dropped $2.5T$ meters in $T$ seconds. The anchor's elevation relative to the water's surface can be found by taking the height from which the anchor was released and subtracting the distance it dropped. We can express this with the equation $E=H-2.5T$, where: $E$ represents the anchor's elevation relative to the water's surface at a given time (in meters) $H$ represents the height from which the anchor was released (in meters) $T$ represents the time (in seconds) We want to find $H$, so let's first solve the equation for $H$ : $ \begin{aligned}E&=H-2.5T\\ H&=E+2.5T\end{aligned}$ Now, we know that after $45$ seconds $(T={45})$, the anchor was $40$ meters below the water's surface $(E={-40})$. Let's plug these values into the equation to find the value of $H$. $ H={-40}+2.5\cdot{45}=72.5$ Therefore, the anchor was released from a height of $72.5$ meters above the water's surface. To find how long it took the anchor to reach the water's surface, we can plug $E=0$ into the equation and solve for $T$. $ \begin{aligned}72.5&=0+2.5T\\ 2.5T&=72.5\\ T&=29\end{aligned}$ The anchor was released from a height of $72.5$ meters above the water's surface. It took the anchor $29$ seconds to reach the water's surface.